JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A cord is wound round the circumference of wheel of radius \(r\). The axis of the wheel is horizontal and the moment of inertia about it is \(I. \,A\) weight \(mg\) is attached to the cord at the end. The weight falls from rest. After falling through a distance \( 'h '\), the square of angular velocity of wheel will be ..... .
- A \(\frac{2 mgh }{ I +2 mr ^{2}}\)
- B \(\frac{2 mgh }{ I + mr ^{2}}\)
- C \(2 gh\)
- D \(\frac{2 gh }{ I + mr ^{2}}\)
Answer & Solution
Correct Answer
(B) \(\frac{2 mgh }{ I + mr ^{2}}\)
Step-by-step Solution
Detailed explanation
\(mgh =\frac{1}{2} I \omega^{2}+\frac{1}{2} mv ^{2}\) \(v =\omega r\) \(mgh =\frac{1}{2} I \omega^{2}+\frac{1}{2} m \omega^{2} r ^{2}\) \(\frac{2 mgh }{\left( I + mr ^{2}\right)}=\omega^{2}\)
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